Let f(x) be a continuous function such that f(a - x) + f(x) = 0 for all x ∈ [0, a]. Then $\int_0^a\frac{dx}{1+e^{f(x)}}$ is equal to:

$\frac{a}{2}$

f(x) = – f(a – x)

$I=\int\limits_0^a\frac{dx}{1+e^{f(x)}}dx=\int\limits_0^a\frac{dx}{1+e^{f(a-x)}}=\int\limits_0^a\frac{dx}{1+e^{-f(x)}}=\int\limits_0^a\frac{e^{f(x)}}{1+e^{f(x)}}dx$

$⇒I=\frac{1}{2}\int\limits_0^a\frac{1+e^{f(x)}}{1+e^{f(x)}}dx=\frac{1}{2}(a-0)=\frac{a}{2}$